Solutions to $(x+y)f(x+y)=xf(x)+yf(y)$

Question

If $x$ is a differentiable function of $t$ and if we define

$$ f(x)=\frac{x^\prime}{x} $$

then $f$ satisfies logarithmic-like properties

but $f$ also satisfies the non-logarithmic-like property

Are there any algebraic functions satisfying the functional equation $(4)$?

Note: It is fairly easy to show that $f(x)=f(-x)$.

Any additive function divided by identity function will work. — Mohsen Shahriari, Sep 06 '17 at 01:03
@MohsenShahriari Not sure I understand. If you mean something like $f(x)=\frac{\ln x}{x}$ then $xf(x)+yf(y)=xyf(xy)$, not $(x+y)f(x+y)$. — John Wayland Bales, Sep 06 '17 at 01:14
An additive function is $g$ such that $g(x+y) = g(x)+g(y)$. The only continuous ones are $g(x) = k x$, but there are exotic discontinuous (and non-measurable) solutions as well. — Robert Israel, Sep 06 '17 at 01:19
@RobertIsrael Regarding your previous proof that no such function would exist, what had been the error with it? — Andrew Tawfeek, Sep 06 '17 at 01:21
@orangeskid I was supposing $\mathbb{R}$ or even $\mathbb{Z}$. Both deleted answers noted correctly that constant functions are solutions. Perhaps I should have specified 'non-trivial' but those seem to be the only solutions, or at least the only differentiable solutions as noted by@RobertIsrael. — John Wayland Bales, Sep 06 '17 at 01:33

Robert Israel · Accepted Answer · 2017-09-06T01:23:15.277

6

If $g(x) = x f(x)$, the equation (4) says $g(x+y) = g(x) + g(y)$, i.e. $g$ is additive. See Cauchy's functional equation.

edited Sep 06 '17 at 01:23

answered Sep 06 '17 at 01:14

Robert Israel

user577215664 · Answer 2 · 2017-09-06T01:30:20.667

0

$(x+y)f(x+y)=xf(x)+yf(y) $ (4)

Let $y=x$ you get that $f(2x)=f(x) $

Let $y=2x$ you get $f(3x) = f(x) $

Let $y=nx$ you get $f(nx) = f(x) $

And we know that $f(x) = f(-x) $

$f(1)=f(2)=f(4)=f(5)....$

so f is constant function ?

edited Sep 06 '17 at 01:30

answered Sep 06 '17 at 01:19

user577215664

2 Answers2